Limit of Characteristic Function of Rationals I'm trying to show that the limits $ \lim_{x \rightarrow x_0+} \chi_{\mathbb{Q}}(x)  $ and $ \lim_{x \rightarrow x_0-} \chi_{\mathbb{Q}}(x)$ don't exist.  [$\chi_{\mathbb{Q}}(x)$ is $1$ if $x \in \mathbb{Q}$ and $0$ otherwise.] 
My attempt at a proof: Choose $\epsilon = \frac{1}{2}.$ Every neighborhood of $x_0$ contains both rationals and irrationals, so the function takes on both the values $0$ and $1$, so there is no interval such that $|{\chi_{\mathbb{Q}}(x) - L}| < \epsilon \ \ \forall x$. 
I believe that this argument works for both limits, as the function is symmetrical. Is my reasoning correct?
Also, would the same proof work for the characteristic function of the irrationals? I can't think of any reason why it would be different.
 A: Your argument is fine. More generally, using your argument we can prove the following theorem,

Theorem. Let $D\subset \mathbb{R}$ be a dense subset of $\mathbb{R}$ such that $\mathbb{R}\setminus D$ is also an dense subset of $\mathbb{R}$.  Then $\displaystyle\lim_{x\to x_0+} \chi_D(x)$ doesn't exist for all $x_0\in \mathbb{R}$.

I will be rewriting your proof (with slight modifications) for the sake of completeness,

Proof. Choose $\varepsilon=\dfrac{1}{2}$. If for some $L$ we would have $\displaystyle\lim_{x\to x_0+} \chi_D(x)=L$ then this would imply that there exists $\delta>0$ such that for all $x\in (x_0,x_0+\delta)$ we would have $$|\chi_D(x)-L|<\dfrac{1}{2}\tag{1}$$Now since $D$ and $\mathbb{R}\setminus D$ is dense, $(x_0,x_0+\delta)$ contains $x$'s such that $x\in D$ and $y$'s such that $y\in \mathbb{R}\setminus D$. So for some $y\in (x_0,x_0+\delta)$ we have $\chi_D(y)=1$. In this case from $(1)$ we have, $$|1-L|<\dfrac{1}{2}\implies \dfrac{1}{2}<L<\dfrac{3}{2}\tag{2}$$ and for some $z\in (x_0-\delta,x_0+\delta)$ we have $\chi_D(z)=0$. In this case from $(1)$ we have, $$|L|<\dfrac{1}{2}\implies -\dfrac{1}{2}<L<\dfrac{1}{2}\tag{3}$$Now observe that $(2)$ contradicts $(3)$ and with this contradiction we are done.

Remark 1. In a similar manner you can prove that following theorem,

Let $D\subset \mathbb{R}$ be a dense subset of $\mathbb{R}$ such that $\mathbb{R}\setminus D$ is also an dense subset of $\mathbb{R}$.  Then $\displaystyle\lim_{x\to x_0-} \chi_D(x)$ doesn't exist for all $x_0\in \mathbb{R}$.

Remark 2. You can give another proof of the theorem just stated by using sequential definition of the limit of a function. Can you do that?   
